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(B) Let matrix $B = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix}$.Given $AB = BA$,we have:$\begin{bmatrix} 0 & 1 & 0 \ 1 & 0 & 0

Vedclass · Accepted Answer

$3125$

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(B) Let matrix $B = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix}$.Given $AB = BA$,we have:$\begin{bmatrix} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{bmatrix}$Performing matrix multiplication:$\begin{bmatrix} d & e & f \ a & b & c \ g & h & i \end{bmatrix} = \begin{bmatrix} b & a & c \ e & d & f \ h & g & i \end{bmatrix}$Comparing the corresponding elements:$d = b, e = a, f = c, g = h$Thus,matrix $B$ takes the form:$B = \begin{bmatrix} a & b & c \ b & a & c \ g & g & i \end{bmatrix}$Since each entry $a, b, c, g, i$ can be chosen from the set $\{1, 2, 3, 4, 5\}$,there are $5$ choices for each of the $5$ independent variables.Total number of matrices $B = 5 	imes 5 	imes 5 	imes 5 	imes 5 = 5^5 = 3125$.

Let $A = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$. Then the number of $3 \times 3$ matrices $B$ with entries from the set $\{1, 2, 3, 4, 5\}$ and satisfying $AB = BA$ is $....$

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